题解 | #吃瓜群众#

using namespace std;
int main()
{
    int weight;
    int flag = 0;
    cin >> weight;
    int arr[100];
    for(int i = 2;i<weight;i+=2)
    {
        arr[i] = i;
        for(int j = 2;j<weight;j+=2)
        {
            arr[j] = j;
            if(arr[i]+arr[j]==weight)
            {
                cout << "YES, you can divide the watermelon into two even parts.";
                flag=1;
                break;
            }
        }
        if(flag==1)
            break;
    }
    if(flag==0)
        cout << "NO, you can't divide the watermelon into two even parts.";
    return 0;
}

我居然想的是遍历数组去相加能不能等于weight，还搞了个时间复杂度为n^2的方法啊！打开评论区看到你们那么简单的方法破防了哈哈哈，向各位学习！